#import "@preview/physica:0.9.6": *

= problem 2

== a)

$h(x) := f(x) g(x)$ is odd for odd $f$ and even $g$, since
$
  h(x) = -f(-x) g(-x) = - (f(-x) g(-x)) = -h(-x)
$

== b)

if $f$ and $g$ are both odd or even, then $h(x) := f(x) g(x)$ is even

1. for the first case: $f$ and $g$ are both even
  $
    h(x) = f(x) g(x) = f(-x) g(-x) = h(-x)
  $

2. for the second case: $f$ and $g$ are both odd
  $
    h(x) & = f(x) g(x) \
         & = (-f(-x)) (-g(-x)) = f(-x) g(-x) = h(-x)
  $

== c)

if $f$ is odd and $g$ is even then both $f compose g$ and $g compose f$ are
even, since
1. $f(g(x)) = f(g(-x))$, since g is even
2. $g(f(x)) = g(-f(-x)) = g(f(-x))$

== d)

if $f$ is odd and $L > 0$,
$
  integral_(-L)^L f(x) dd(x) & = integral_(-L)^0 f(x) dd(x)
                               + integral_0^L f(x) dd(x) \
                             & = integral_0^L f(-x) (-1) dd(x)
                               + integral_0^L f(x) dd(x) \
                             & = -integral_0^L f(-x) dd(x)
                               + integral_0^L f(x) dd(x) \
                             & = -integral_0^L f(x) dd(x)
                               + integral_0^L f(x) dd(x) \
                             & = 0
$

== e)

if $f$ is even and $L > 0$,
$
  integral_(-L)^L f(x) dd(x) & = integral_(-L)^0 f(x) dd(x)
                               + integral_0^L f(x) dd(x) \
                             & = integral_0^L f(-x) dd(x)
                               + integral_0^L f(x) dd(x) \
                             & = integral_0^L f(x) dd(x)
                               + integral_0^L f(x) dd(x) \
                             & = 2 integral_0^L f(x) dd(x)
$